# Definite Integral in u and u'

Suppose I want to solve $$\int_{-1}^1 u(x)^{\ 2}\ dx$$. I should be able to write this as $$\int_{-1}^1 u^{\ 2} \frac{1}{u'}\ du$$.

Since $$\int_{-1}^1 u'\ du$$ is presumably just $$u(1) - u(-1)$$, can I also find the definite integral of an integrand in $$u$$ and $$u'$$, such as $$\int_{-1}^1 u^{\ 2} \frac{1}{u'}\ du$$? How would I proceed in doing so?

• you can substitute like that only if $u'$ is constant. otherwise you would still have a function of x in the integral. you also have to interchange limits of integration, so the integral would be from $u(-1)$ to $u(1)$. – Yizhar Amir Mar 26 at 16:43
• Unless $u(1)=1$ and $u(-1)=-1$ and $u(x)$ is increasing in the interval, your substitution won't be valid. – herb steinberg Mar 26 at 17:32
• @YizharAmir Isn't what I did very similar to youtu.be/_60sKaoRmhU?t=840 I just multiplied the $dx$ by $\frac{du}{du}$ to get $\frac{dx}{du}\ du \implies \frac{1}{u'}\ du$. – user10478 Mar 27 at 3:48